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My goal is to compute

\begin{equation} I = \det \left(\mathbf{I} + \mathbf{A}\right) \end{equation} where $\mathbf{A}$ is a $n \times n$ checkerboard matrix filled with Catalan numbers: $$ \left\{ \begin{array}{crc} \mathbf{A}_{ij} = C_{p-1} \alpha^{2(p-1)} &\mbox{ if }& i+j=2p \mbox{, and with }C_p= \frac{1}{p+1}\binom {2p} {p}\\ \mathbf{A}_{ij} = 0 &\mbox{ if }& i+j \mbox{ is odd.} \end{array} \right. $$ with $\alpha >0$ a parameter.

Numerically, it seems that $I$ has a limit when $n\to +\infty$ if $\alpha<1/2$ and diverges to $\infty$ otherwise.

Any idea ?

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  • $\begingroup$ What are the first few values? Anything in the Online Encyclopedia of Integer Sequences? $\endgroup$ Apr 21, 2015 at 12:54
  • $\begingroup$ what relation does $p$ have to $n$? $\endgroup$
    – Suvrit
    Apr 21, 2015 at 13:16
  • $\begingroup$ @Suvrit :p is just (i+j)/2 when i+j is even $\endgroup$
    – user16215
    Apr 21, 2015 at 13:51
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    $\begingroup$ Maybe some ideas in cs.ucsb.edu/~omer/DOWNLOADABLE/catalan_hankel09.pdf may be helpful. or also in: arxiv.org/pdf/0709.3044.pdf $\endgroup$
    – Suvrit
    Apr 21, 2015 at 14:30
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    $\begingroup$ Note that there is no interaction between the even and odd coordinates in your matrix. In other words, if you rearrange rows and columns to put the even-numbered rows first and then the odd-numbered ones, you get a block-diagonal matrix and you should probably be looking at the blocks individually. It's also unclear at what numbers your matrix starts, you should write down the top left corner explicitly. There's a nice combinatorial interpretation for these; see jacobi.math.wvu.edu/~jerzy/research/21catalan.pdf. $\endgroup$ Apr 21, 2015 at 14:44

2 Answers 2

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If you consider the operator on functions that sends $f(x)$ to

$$Tf(y) = \int_{-2}^2 \frac{ \sqrt{4-x^2} }{2\pi } \frac{1}{ 1- \alpha x y} f(\alpha x) dx$$

then this is a well-defined integral operator from functions on $[-2\alpha,2\alpha]$ to functions on $[-2\alpha, 2\alpha]$ as long as $\alpha<1/2$.

Moreover, if $f(x)=x^{i-1}$, then the coefficient of $y^{j-1}$ in $Tf(y)$ is exactly $\mathbf A_{ij}$. So the limit of your determinant should be exactly the Fredholm determinant of this integral operator.

I'm not sure if this helps at all.

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Since $$\binom{2p}{p} =O\left(\frac{4^p}{\sqrt{p}}\right),$$ the coefficients of your matrix are unbounded when $\alpha > 1/2,$ while for $\alpha < 1/2$ the matrix $A$ is well-approximated by its piece where $i+j$ is small, so your observations are not surprising. The interesting case is $\alpha = 1/2,$ where I have no intuition.

EDIT A mathematica experiment seems to indicate that the determinants converge slowly for $\alpha = 1/2.$

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  • $\begingroup$ Thank you! Just a minor comment : the fact that the coefficient are unbounded does not imply that the determinant will be large, I think. $\endgroup$
    – user16215
    Apr 21, 2015 at 19:35
  • $\begingroup$ @user16215 It does not logically imply it, it suggests it - notice that I never claimed that this was proof. $\endgroup$
    – Igor Rivin
    Apr 21, 2015 at 19:41

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